OFFSET
1,5
COMMENTS
Informally: starting from k=n, keep on replacing p^e, the maximal power of the smallest odd prime factor in k, with (1 + p + p^2 + ... + p^e), until a power of 2 is reached. Sequence counts the steps needed.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..65537
FORMULA
EXAMPLE
For n = 15 = 3*5, we obtain the following path, when starting from k = n, and when we always replace the maximal power of the lowest odd prime factor, p^e of k with sigma(p^e) = (1 + p + p^2 + ... + p^e) in the prime factorization k: 3^1 * 5^1 -> (1+3)*5 = 20 = 2^2 * 5 -> 4 * (1+5) = 24 = 2^3 * 3^1 -> 2^3 * 2^2 = 2^5, thus it took three iterations to reach a power of two, and a(15) = 3.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 30 2020
STATUS
approved